

SSCommitted
Group: General Forum Members
Last Login: Tuesday, May 29, 2012 11:22 AM
Points: 1,755,
Visits: 4,652


Thanks Frédéric :)
I dare say that's my first ever 'publication' in a foreign language  great stuff!
Ryan Randall
Solutions are easy. Understanding the problem, now, that's the hard part.




Grasshopper
Group: General Forum Members
Last Login: Saturday, September 19, 2015 7:50 AM
Points: 15,
Visits: 127


I hope it won't be the last !
A +




Forum Newbie
Group: General Forum Members
Last Login: Tuesday, July 12, 2016 5:14 PM
Points: 2,
Visits: 61


regarding the original query. It seems repetition of the characters is not allowed. This is therefore not a permutations calculator but is a combinations calculator.




Forum Newbie
Group: General Forum Members
Last Login: Saturday, July 2, 2016 11:43 AM
Points: 5,
Visits: 187


This uses a binary mask to select the unique sets where one of each character are present. It runs in about 14 seconds.
declare @t varchar(10) = 'ABCDEFGH'
;with src(t,n,p) as ( select substring(@t,1,1),1,power(2,0) union all select substring(@t,n+1,1),n+1,power(2,n) from src where n < len(@t) ) select s1.t+s2.t+s3.t+s4.t+s5.t+s6.t+s7.t+s8.t from src s1, src s2, src s3, src s4, src s5, src s6, src s7, src s8 where s1.p+s2.p+s3.p+s4.p+s5.p+s6.p+s7.p+s8.p=power(2,len(@t))1




SSCForever
Group: General Forum Members
Last Login: Yesterday @ 9:11 PM
Points: 41,438,
Visits: 38,772


Frédéric BROUARD (1/8/2009) This calculus can be done in only one query(I think I am the first to demontsrate how to do that in one query only !) ;)  lest's assume that this table containes all datas to be permuted : CREATE TABLE T_CMB (CMB_DATA VARCHAR(8))  let's assume that the joker character ; (dot comma) is not used inside the data : INSERT INTO T_CMB VALUES ('ABC') INSERT INTO T_CMB VALUES ('DEF') INSERT INTO T_CMB VALUES ('GHI')  the following query does the permutations WITH T_DATA AS (SELECT CMB_DATA, 1 AS COMBINAISON, ROW_NUMBER() OVER(ORDER BY CMB_DATA) AS ORDRE, COUNT(*) OVER() AS N FROM T_CMB), T_RECUR AS (SELECT CAST(CMB_DATA AS VARCHAR(max)) +';' AS CMB_DATA, COMBINAISON, ORDRE, N FROM T_DATA UNION ALL SELECT T1.CMB_DATA + ';' + T2.CMB_DATA, T2.COMBINAISON + 1, ROW_NUMBER() OVER(PARTITION BY T1.COMBINAISON ORDER BY T2.CMB_DATA) ORDRE, T1.N FROM T_DATA AS T1 CROSS JOIN T_RECUR AS T2 WHERE T2.COMBINAISON < T1.N  this line must be delete if you want a repetitive permutation AND T2.CMB_DATA NOT LIKE '%' + T1.CMB_DATA +';%' ), T_COMBINE AS (SELECT CMB_DATA, ROW_NUMBER() OVER(ORDER BY CMB_DATA) AS ORDRE FROM T_RECUR WHERE COMBINAISON = N), T_N AS (SELECT 1 AS N UNION ALL SELECT N + 1 FROM T_N WHERE N + 1 <= ALL (SELECT LEN(CMB_DATA) FROM T_COMBINE)), T_SOL AS (SELECT *, REVERSE(SUBSTRING(CMB_DATA, 1, N1)) AS SOUS_CHAINE, REVERSE(SUBSTRING(REVERSE(SUBSTRING(CMB_DATA, 1, N1)), 1, CASE WHEN CHARINDEX(';', REVERSE(SUBSTRING(CMB_DATA, 1, N1)))  1 = 1 THEN LEN(CMB_DATA) ELSE CHARINDEX(';', REVERSE(SUBSTRING(CMB_DATA, 1, N1)))  1 END)) AS DATA FROM T_COMBINE INNER JOIN T_N ON SUBSTRING(CMB_DATA, N, 1) = ';') SELECT DATA AS CMB_DATA, ORDRE AS PERMUTATION FROM T_SOL CMB_DATA PERMUTATION   ABC 1 DEF 1 GHI 1
ABC 2 GHI 2 DEF 2
DEF 3 ABC 3 GHI 3
DEF 4 GHI 4 ABC 4
GHI 5 ABC 5 DEF 5
GHI 6 DEF 6 ABC 6 If you want a permutation with repetitive datas, simply delete the 18e line : AND T2.CMB_DATA NOT LIKE '%' + T1.CMB_DATA +';%' You'll get : CMB_DATA PERMUTATION   ABC 1 ABC 1 ABC 1
ABC 2 ABC 2 DEF 2
ABC 3 ABC 3 GHI 3
ABC 4 DEF 4 ABC 4
ABC 5 DEF 5 DEF 5
ABC 6 DEF 6 GHI 6
ABC 7 GHI 7 ABC 7
ABC 8 GHI 8 DEF 8
ABC 9 GHI 9 GHI 9
DEF 10 ABC 10 ABC 10
DEF 11 ABC 11 DEF 11
DEF 12 ABC 12 GHI 12
DEF 13 DEF 13 ABC 13
DEF 14 DEF 14 DEF 14
DEF 15 DEF 15 GHI 15
DEF 16 GHI 16 ABC 16
DEF 17 GHI 17 DEF 17
DEF 18 GHI 18 GHI 18
GHI 19 ABC 19 ABC 19
GHI 20 ABC 20 DEF 20
GHI 21 ABC 21 GHI 21
GHI 22 DEF 22 ABC 22
GHI 23 DEF 23 DEF 23
GHI 24 DEF 24 GHI 24
GHI 25 GHI 25 ABC 25
GHI 26 GHI 26 DEF 26
GHI 27 GHI 27 GHI 27 The french version is on my blog : http://blog.developpez.com/sqlpro?title=calculs_de_tous_les_arrangements_mathemaCU  Frédéric BROUARD, Spécialiste modélisation, bases de données, optimisation, langage SQL. Le site sur le langage SQL et les S.G.B.D. relationnels : http://sqlpro.developpez.com/Expert SQL Server http://www.sqlspot.com : audit, optimisation, tuning, formation * * * * * Enseignant au CNAM PACA et à l'ISEN à Toulon * * * * *
At a "9" count, you're also the first to get beat by the While loop by a factor of more than 30 even with discarded results enabled.
Jeff Moden "RBAR is pronounced "reebar" and is a "Modenism" for "RowByAgonizingRow".
First step towards the paradigm shift of writing Set Based code: Stop thinking about what you want to do to a row... think, instead, of what you want to do to a column."
Helpful Links: How to post code problems How to post performance problems




Forum Newbie
Group: General Forum Members
Last Login: Saturday, July 2, 2016 11:43 AM
Points: 5,
Visits: 187


I have a new solution. This one is Recursive CTE but only builds valid solutions instead of all solutions. It will produce the 3.6 million permutations for a 10 character string in 3:02 minutes.
declare @t varchar(10) = 'ABCDEFGHIJ'
;with s(t,n) as ( select substring(@t,1,1),1 union all select substring(@t,n+1,1),n+1 from s where n<len(@t) ) ,j(t) as ( select cast(t as varchar(10)) from s union all select cast(j.t+s.t as varchar(10)) from j,s where patindex('%'+s.t+'%',j.t)=0 ) select t from j where len(t)=len(@t)



